show $\lim_{x\rightarrow\infty} x^n\exp(-x^2)=0$ I want to show for all $n\in\mathbb N$ 
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$
I am pretty sure that I have to use L'Hospital. I've tried induction:
$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$
And for $n\rightarrow n+1$:
$$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$
And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.
Any hints?
 A: From where you left it:
$$\lim_{x\to\infty}\frac{(n+1)\color{red}{x^n}}{2\color{red} xe^{x^2}}=\frac{n+1}2\lim_{x\to\infty}\frac{x^{n-1}}{e^{x^2}}\stackrel{\text{Inductive Hyp.}}=\frac{n+1}2\cdot 0=0$$
A: For $ \ x > 1 \ , $
$$ 0 \ \le \ e^{-x^2} \ \le \ e^{-x} $$ 
and also $ \ 0 \ < \ x^n \ \text{for} \ n  \in \mathbb{N}$ , so
$$ 0 \ \le \ x^ne^{-x^2} \ \le \ x^ne^{-x} \ .$$
It is much easier to show $ \ \lim_{x \rightarrow \infty} \  x^ne^{-x} \ = \ 0  \ $ through l'Hopital's Rule.  Thence, the "Squeeze Theorem" leads to
$$ 0 \ \le \ \lim_{x \rightarrow \infty} x^ne^{-x^2} \ \le \ \lim_{x \rightarrow \infty} x^ne^{-x} \ \Rightarrow \ \lim_{x \rightarrow \infty} x^ne^{-x^2} \ = \ 0 \ . $$
A: $$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=\left( \lim_{x\rightarrow\infty}\frac{x}{\exp(x^2/n)}\right)^n=0$$
Now, you only need to apply L'H once.
Second solution
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{\exp(n \ln x)}{\exp(x^2)}=\lim_{x\rightarrow\infty}\exp(n \ln x-x^2)=0$$
A: Note that for positive $x$, $x^{2n}\le e^{x^2}$ (by considering the taylor series), so $\frac{x^n}{e^{x^2}}\le\frac{x^n}{x^{2n}}=\frac{1}{x^n}$, hence $\lim\limits_{x\to\infty}\frac{x^n}{e^{x^2}}\le\lim\limits_{x\to\infty}\frac{1}{x^n}=0$.
A: Applying L'Hospital $n$ times would do the trick too (assuming $n$ is fixed):
$$
\lim_{x\to\infty}\frac{x^{n}}{e^{x^2}} = \lim_{x\to\infty}\frac{(x^{n})'}{(e^{x^2})'} = \lim_{x\to\infty}\frac{nx^{n-1}}{e^{x^2} 2x},
$$
again we have an $\infty/\infty$ type limit, repeating L'Hospital until we get a constant in the denominator:
$$
\lim_{x\to\infty}\frac{(x^{n})^{(n)}}{(e^{x^2})^{(n)}} =  \lim_{x\to\infty}\frac{n!}{e^{x^2} p(x)} = 0,
$$
where $p(x)$ is a degree $n$ polynomial in $x$.
